Motion in A Straight Line Test

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Motion in A Straight Line Test - 22

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Question 1
1 / -0

A body thrown up with a velocity of 98 m/s reaches a point P in its path 7 second after projection. Since its projection it comes back to the same position after:

A
13s

B
14s

C
6s

D
22s

Solution

Here, $$u=98m/s$$, $${t}_{1}=7s$$. When body reaches maximum height, its velocity becomes $${v}_{1}=0$$.
Maximum height reached can be obtained using:
$${v}^{2}={u}^{2}+2ah$$
$$h_{max}=\dfrac{{u}^{2}}{2g}$$
Now considering initial velocity as $${v}_{1}=0$$, displacement as $$s=\dfrac{{u}^{2}}{2g}$$,
We can find time for the body to come back from the maximum height using:
$$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }\\ \dfrac { { u }^{ 2 } }{ 2g } =\dfrac { 1 }{ 2 } g{ t }^{ 2 }\\ t=\dfrac { u }{ g } \\ $$
Thus, total time, $$t=\dfrac{2u}{g}=20\ s$$
Now the time it takes to reach point P from ground and to reach from point P to ground is same i.e. $$7s$$ each time.
Thus, it takes $$20s - 7s = 13s$$ to come back to the same point P since its projection.
Question 2
1 / -0

A particle is projected vertically up and another is let fall to meet at the same instant. If they have velocities equal in magnitude when they meet, the distances travelled by them are in the ratio

A
1:1

B
1:2

C
3:1

D
2:2

Solution

$$\text{Let}\ A\ \& B\ \text{meet at P.}$$
$$\text{For}\ A \Rightarrow v = gt .................. (1)$$
$$\text{For}\ B \Rightarrow v^{'}= u - gt ....................(2)$$
$$\text{The magnitude of } v\ \text{and}\ v^{'}\ \text{to be same}$$
$$|v| = |v^{'}|$$
$$\Rightarrow u - gt = gt$$
$$u = 2gt \Rightarrow t = \dfrac {u}{2g}$$
$$\text{So, dist. travelled by B:}$$
$$\Rightarrow S = ut - \dfrac {1}{2}gt^2$$
$$S= u\left(\dfrac {u}{2g}\right) - \dfrac {1}{2}g\left(\dfrac {u}{2g}\right)^2$$

$$S=\dfrac{u^2}{2g}-\dfrac{u^2}{8g}= \dfrac {3}{8}\left(\dfrac {u^2}{g}\right)$$
$$\text{Dist. travelled by}\ B\Rightarrow S^{'} = \dfrac {1}{2}gt^2 = \dfrac {1}{2}g\left(\dfrac {u}{2g}\right)^2 = \dfrac {u^2}{8g}$$
So, $$ \dfrac {S}{S^{'}} = \dfrac {3}{1}$$
Question 3
1 / -0

A ball of mass $$100$$ gm is projected vertically upwards from the ground with a velocity of $$49$$ $$m/s$$. At the same time another identical ball is dropped from a height of $$98$$ m. After some time the two bodies collide. When they collide, their velocities are (Given $$g=9.8\; m/s^2$$)

A
29.4 m/s upwards, 29.4 m/s downwards

B
29.4 m/s upwards, 19.6 m/s downwards

C
19.6 m/s upwards, 19.6 m/s downwards

D
None

Solution

Equations to use :
time of meet $$t=\frac{h}{u}$$
$$v_{1}=gt$$ ( for freely falling body)
$$v_{2}=u-gt$$ ( for body projected up)

For A: $$s_A = 49t - \dfrac {1}{2}gt^2 - (1)$$
For B: $$s_B = \dfrac {1}{2}gt^2 - (2)$$
$$s_A + s_B = 98 = 49t \ \ \ \ (from (1) \& (2))$$
$$\Rightarrow t = 2 sec$$

So, $$v_A = u_A - gt = 49 - 9.8 \times 2 = 29.4\ m/s $$ upwards
$$v_B = gt = 9.8 \times 2 = 19.6\ m/s $$ downwards
Question 4
1 / -0

A body projected up with a speed $$u$$ took $$T$$ seconds to reach the maximum height $$H$$. Pick out the correct statement

A
It reaches $$\dfrac{H}{2}$$ in $$\dfrac {T}{2}$$s

B
It acquires velocity $$\dfrac{u}{2}$$ in $$\dfrac{T}{2}$$s

C
Its velocity is $$\dfrac{u}{2}$$ at $$\dfrac{H}{2}$$

D
Same velocity at $$2T$$s

Solution

$$v = u - gT$$
$$\Rightarrow T = \dfrac {u}{g} \ \ \ .... (1) $$
$$H = uT - \dfrac {1}{2}gT^2$$
In $$ \dfrac {T}{2} sec \Rightarrow h^1 = u (\dfrac {T}{2}) - \dfrac {1}{2}g \dfrac {T^2}{4}$$

Hence $$h^1 \neq \dfrac {H}{2}$$
$$v = u -gt$$
At, $$ t = \dfrac {T}{2}$$
$$v = u - g\dfrac {T}{2} = u - \dfrac {u}{2} = \dfrac {u}{2} \ \ \ (from (1) T = \dfrac {u}{g})$$
$$v^2 = u^2 - 2gH$$

At $$\dfrac {H}{2}, v^2 = u^2 - 2g\dfrac {H}{2} = u^2 - gH \neq \dfrac {u^2}{4} \Rightarrow v \neq \dfrac {u}{2}$$

Hence, option $$B$$ is correct.
Question 5
1 / -0

From the top of a tower $$36$$ m high, a body is dropped and at the same time another body is projected vertically upward from the ground. If they meet midway, find the initial velocity of the projected body and its velocity when the two bodies come together

A
$$5\sqrt{6}m, 0\ m/s$$

B
$$\dfrac{42}{\sqrt{5}}m,0\ m/s$$

C
$$6\sqrt{6}m , 2\ m/s$$

D
$$8\sqrt{6}m, 4.5\ m/s$$

Solution

Equation to use :
for first body $$x=\displaystyle \frac{1}{2}gt^{2}$$
for second body $$(h-x)=ut-\displaystyle \frac{1}{2}gt^{2}$$

Now solving,

For A $$\Rightarrow 18 = \dfrac {1}{2}gt^2$$
$$\sqrt {\dfrac {2 \times 18}{g}} = t_1$$

For B $$\Rightarrow 18 = ut - \dfrac {1}{2}gt^2 (t = t_1)$$
$$\Rightarrow 18 = u \sqrt {\dfrac {36}{g}} - \dfrac {g}{2} \times \dfrac {36}{g} \Rightarrow 36 = u \sqrt {\dfrac {36}{g}}$$
$$\Rightarrow u = 18.78 = \dfrac {42}{\sqrt 5} m/s$$

When they come in contact:
Velocity of B is, $$v_B = u - gt$$ $$ = \dfrac {42}{55} - 9.8 \times 1.916$$
$$v_B = 0\ m/s$$
Question 6
1 / -0

A rocket is fired and ascends with constant vertical acceleration of $$10m/s^{2}$$ for 1 minute. Its fuel is exhausted and it continues as a free particle.The maximum altitude reached is ($$g=10m/s^{2}$$)

A
18 km

B
36 km

C
72 km

D
108km

Solution

For $$1$$ minute, distance, $$d$$ travelled is:
$$d = ht^0 + \dfrac {1}{2} at^2 = \dfrac {1}{2} \times 10 \times (60)^2 = \dfrac {36000}{2} = 18000\ mt$$
$$v = u + at = 10 \times 60 = 600\ m/s$$

Now, for max $$m$$ height:
$$v_1^2 = 4_1^2 - 2gs (v_1 = v)$$
$$\Rightarrow s = \dfrac {(600)^2}{2g} = \dfrac {600 \times 600}{2 \times 10} = 18000 \ mt$$
$$\Rightarrow $$ Total attitude $$= d + s = 18000 + 18000$$
$$= 36000\ m$$
$$= 36\ Km$$
Question 7
1 / -0

A stone projected vertically up from the ground reaches a height $$y$$ in its path at $$t_{1}$$ seconds and after further $$t_{2}$$ seconds reaches the ground. The height $$y$$ is equal to

A
$$\frac{1}{2}g(t_{1}+t_{2})$$

B
$$\frac{1}{2}g(t_{1}+t_{2})^{2}$$

C
$$\frac{1}{2}gt_{1}t_{2}$$

D
$$g t_{1}t_{2}$$

Solution

$$y = ut_1 - \dfrac {1}{2}gt_1^2$$
$$\dfrac {t_1 + t_2}{\Downarrow} = \dfrac {2u}{g}$$
(total time of journey)
So, $$u = g\dfrac {(t_1 + t_2)}{2}$$
$$\Rightarrow y = g\dfrac {(t_1 + t_2)}{2}t_1 - \dfrac {1}{2}gt_1^2 = \dfrac {1}{2}gt_1t_2$$
Question 8
1 / -0

A paper weight is dropped from the roof of a block of multistorey flats, each storey being $$3$$ meters high. It passes the ceiling of the $$20^{th}$$ storey at $$30$$ $$m/s$$. If $$g = 10 m/ s^{2}$$, how many storey does the flat have?

A
25

B
30

C
35

D
40

Solution

$$v^2 = u^2 + 2gs$$
$$\Rightarrow (30)^2 = 0^2 + 2 \times 10 \times s$$

$$s = \dfrac {900}{20} = 45\ m$$
$$\Rightarrow $$ It has travelled $$\dfrac {45}{3} = 15 $$ storeys
And it is $$21^{st}$$ storey from bottom.

Hence, total storeys $$ = 20 + 15 = 35 $$ storeys.
Question 9
1 / -0

A stone is dropped from a height $$h$$. Simultaneously another stone is thrown up from the ground which reaches the height $$4h$$. The two stones cross each other after a time:-

A
$$\sqrt{\dfrac{h}{2g}}$$

B
$$\sqrt{\dfrac{h}{8g}}$$

C
$$\sqrt{8hg}$$

D
$$\sqrt{2hg}$$

Solution

For B: $$u$$ is such that stone B reaches uh.
0 at $$max^m$$ pt

Hence, $$ v^2 = u^2 - 2g_B$$
$$u^2 = 2g (uh) \Rightarrow u = \sqrt {8gh} \ \ \ .....(1)$$

Now $$eq^n $$ for B: $$S_B = ut - \dfrac {1}{2} gt^2 \ \ \ ..... (2)$$

$$eq^n$$ for A: $$S_A = \dfrac {1}{2} gt^2 \ \ \ ..... (3)$$
$$S_A + S_B = h = ut \Rightarrow t = \dfrac {h}{u}$$
From (1), $$u = \sqrt {8gh} \Rightarrow t = \sqrt {\dfrac {h}{8g}}$$
Question 10
1 / -0

A body is thrown vertically up to reach its maximum height in $$t$$ seconds. The total time from the time of projection to reach a point at half of its maximum height while returning ( in seconds ) is

A
$$\sqrt{2}t $$

B
$$\left [ 1+\dfrac{1}{\sqrt{2}} \right ]t$$

C
$$\dfrac{3t}{2}$$

D
$$\dfrac{t}{\sqrt{2}}$$

Solution

At $$max^m$$ height $$ v =o$$,
$$v = u - gt, \Rightarrow t = \dfrac {u}{g} = \sqrt {\dfrac {2H}{g}} - (1)$$
Now situation is,
We want time to reach C.
Time from $$ A to B = t$$
Time from $$ B to C \Rightarrow v_c^2 = v_B^2 + 2g \dfrac {H}{2}$$
$$\dfrac {H}{2} = + \dfrac {1}{2} gt_1^2$$ $$v_C = \sqrt {gH}$$
$$t_1 = \sqrt {\dfrac {H}{g}} \Rightarrow t_1 = \dfrac {t}{\sqrt 2} (from (1))$$
So, total time $$= t + t_1 = t + \dfrac {t}{\sqrt 2} = t (1 + \dfrac {1}{\sqrt 2})$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 22

Result

Motion in A Straight Line Test - 22

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Rank

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SHARING IS CARING

Question 1

13s

14s

6s

22s

Solution

Question 2

1:1

1:2

3:1

2:2

Solution

Question 3

29.4 m/s upwards, 29.4 m/s downwards

29.4 m/s upwards, 19.6 m/s downwards

19.6 m/s upwards, 19.6 m/s downwards

None

Solution

Question 4

It reaches $$\dfrac{H}{2}$$ in $$\dfrac {T}{2}$$s

It acquires velocity $$\dfrac{u}{2}$$ in $$\dfrac{T}{2}$$s

Its velocity is $$\dfrac{u}{2}$$ at $$\dfrac{H}{2}$$

Same velocity at $$2T$$s

Solution

Question 5

$$5\sqrt{6}m, 0\ m/s$$

$$\dfrac{42}{\sqrt{5}}m,0\ m/s$$

$$6\sqrt{6}m , 2\ m/s$$

$$8\sqrt{6}m, 4.5\ m/s$$

Solution

Question 6

18 km

36 km

72 km

108km

Solution

Question 7

$$\frac{1}{2}g(t_{1}+t_{2})$$

$$\frac{1}{2}g(t_{1}+t_{2})^{2}$$

$$\frac{1}{2}gt_{1}t_{2}$$

$$g t_{1}t_{2}$$

Solution

Question 8

25

30

35

40

Solution

Question 9

$$\sqrt{\dfrac{h}{2g}}$$

$$\sqrt{\dfrac{h}{8g}}$$

$$\sqrt{8hg}$$

$$\sqrt{2hg}$$

Solution

Question 10

$$\sqrt{2}t $$

$$\left [ 1+\dfrac{1}{\sqrt{2}} \right ]t$$

$$\dfrac{3t}{2}$$

$$\dfrac{t}{\sqrt{2}}$$

Solution

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